2. A student was given a task to determine the percentage purity of a sample of salicylic acid. The method used by the student to prepare a solution of salicylic acid is described below.
•********0.500 g of an impure sample of salicylic acid was placed in a weighing bottle.
•********The contents were tipped into a beaker and 100 cm3 of distilled water were added.
•********Salicylic acid does not dissolve well in cold water so the beaker and its contents were heated gently until all the solid had dissolved.
•********The solution was poured into a 250 cm3 graduated flask and made up to the mark with distilled water.
b) ****The pH of this solution was measured and a value of 2.50 was obtained.
Calculate the concentration of salicylic acid in this solution.
Assume that salicylic acid is the only acid in this solution. The Ka for salicylic acid is 1.07 × 10–3 mol dm–3. You may represent salicylic acid as HA.
I worked out the answer for this question which was 9.55 mol dm-3.
c) ****Use your answer to part (b) to calculate the mass of salicylic acid (Mr = 138.0) present in the original sample.
I have checked the markscheme and it says to:
(b) × 138.0 / 4
Why does 138 have to be divided by 4? Can someone explain it thank you?
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Chemistry A2 Acids and Bases question HELP!!!!!!!!!!!!!!!!!!!!! watch
- Thread Starter
- 04-11-2017 13:15
- Official Rep
- 06-11-2017 16:16
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- 09-11-2017 21:45
Your [HA] should have been 0.00935 using ka=[H+]^2 ÷ [HA]
Then use c = n ÷ v
To get a number of mols of 2.336x10^-3
Then use n = mass ÷ Mr
Mr = 138
n = 2.336x10-3
To get a mass of 0.322.
I don't understand why the mark scheme divided by 4 but you can get the answer using this method.
- 22-12-2017 23:26
Your [HA] should have been 0.00935 using ka=[H ]^2 / [HA]
C= 0.00935 mol dm-3
V= 250×10-3 dm-3 = 0.250dm3 = ¼ dm3
n = CV= m/Mr
m= CV x Mr = 0.00935 x ¼ x 138= 0.322g
Dr Youssef El AzizLast edited by Dr E. A Youssef; 22-12-2017 at 23:31.